Problem: Given that the diagonals of a rhombus are always perpendicular bisectors of each other, what is the area of a rhombus with side length $\sqrt{89}$ units and diagonals that differ by 6 units?
Because the diagonals of a rhombus are perpendicular bisectors of each other, they divide the rhombus into four congruent right triangles.  Let $x$ be half of the length of the shorter diagonal of the rhombus.  Then $x+3$ is half of the length of the longer diagonal.  Also, $x$ and $x+3$ are the lengths of the legs of each of the right triangles.  By the Pythagorean theorem,  \[
x^2+(x+3)^2=\left(\sqrt{89}\right)^2.
\] Expanding $(x+3)^2$ as $x^2+6x+9$ and moving every term to the left-hand side, the equation simplifies to $2x^2+6x-80=0$.  The expression $2x^2+6x-80$ factors as $2(x-5)(x+8)$, so we find $x=5$ and $x=-8$.  Discarding the negative solution, we calculate the area of the rhombus by multiplying the area of one of the right triangles by 4. The area of the rhombus is $4\cdot\left(\frac{1}{2}\cdot 5(5+3)\right)=\boxed{80}$ square units.

[asy]
unitsize(3mm);
defaultpen(linewidth(0.7pt)+fontsize(11pt));
dotfactor=3;
pair A=(8,0), B=(0,5), C=(-8,0), D=(0,-5), Ep = (0,0);
draw(A--B--C--D--cycle);
draw(A--C);
draw(B--D);
label("$x$",midpoint(Ep--B),W);
label("$x+3$",midpoint(Ep--A),S);
label("$\sqrt{89}$",midpoint(A--B),NE);[/asy]